For the complex ${\text{[Fe(H}}_2{\text{O)}}_5{\text{NO]SO}}_4$, the magnetic moment is found to be $3.87\text{ BM}$. What is the oxidation state of iron and the nature of the NO ligand?

Solution

1. The magnetic moment $$\begin{array}{rl}\mu & =\sqrt{n(n+2)}\\ & =3.87\text{ BM}\end{array}$$ implies $n=3$ unpaired electrons.
2. The complex is the brown ring complex where $\text{Fe}$ is in the $+1$ oxidation state.
3. The configuration of ${\text{Fe}}^{+}$ is $3d^7$.
4. With ${\text{NO}}^{+}$ acting as a strong field ligand, the electrons arrange to give three unpaired electrons.
5. Hence the answer is (B).